Answer:
See below for answers and explanations
Step-by-step explanation:
Problem 1
To convert from polar to Cartesian coordinates, recall that [tex](x,y)=(r\:cos\theta,r\:sin\theta)[/tex], thus, [tex](x,y)=(6\:cos\frac{5\pi}{6},6sin\frac{5\pi}{6})=(6(-\frac{\sqrt{3}}{2}),6(\frac{1}{2}))=(-3\sqrt{3},3)[/tex]
Therefore, C is the correct answer
Problem 2
Convert from polar to Cartesian coordinates:
[tex](x,y)=(-5\:cos\frac{2\pi}{3},-5\:sin\frac{2\pi}{3})=(-5(-\frac{1}{2}),-5(\frac{\sqrt{3}}{2}))=(\frac{5}{2},\frac{-5\sqrt{3}}{2})[/tex]
Therefore, U is the correct answer
Problem 3
Since [tex]r=\sqrt{x^2+y^2}[/tex], then [tex]r^2=x^2+y^2[/tex]. Don't also forget to use the substitution [tex]x=r\:cos\theta[/tex]:
[tex]x^2+y^2-7x=0\\\\r^2-7rcos\theta=0\\\\r(r-7cos\theta)=0\\\\r-7cos\theta=0\\\\r=7cos\theta[/tex]
Therefore, B is the correct answer
Problem 4
[tex]r=(tan\theta)(sec\theta)\\\\r=(\frac{sin\theta}{cos\theta})(\frac{1}{cos\theta})\\ \\\frac{r^2}{r}=(\frac{rsin\theta}{rcos\theta})(\frac{r}{rcos\theta})\\ \\\frac{x^2+y^2}{\sqrt{x^2+y^2}}=(\frac{y}{x})(\frac{\sqrt{x^2+y^2}}{x})\\ \\x^2+y^2=\frac{y(x^2+y^2)}{x^2}\\ \\1=\frac{y}{x^2}\\\\x^2=y[/tex]
Therefore, A is the correct answer
